(1) would like to show that the geometric random variable is memoryless.
Let $ X \sim \mathrm{Geometric}(p)$ show that
$P(X > m+l \mid X > m) = P(X > l)$, for $ m,l \in \{ 1,2,3, \ldots\}$
(2) then I need to show that if X is a random variable $\in \mathbb{N}$ and that $P(X > m+l \mid X > m) = P(X > l)$ is true for X, Then X is geometric distribuered.
Part 1: First I want to find
\begin{align*}
P(X>m) &= \sum\limits_{i=m+1}^\infty P(X= i)\\
&= \sum\limits_{0}^\infty p(1-p)^{i-1} – \sum\limits_{0}^m p(1-p)^{i-1} \\
&= p \sum\limits_{0}^\infty (1-p)^{i-1} -\sum\limits_{0}^m p(1-p)^{i-1} \\
&= \frac{p}{1-(1-p)^2} -\sum\limits_{0}^m p(1-p)^{i-1} \\
\end{align*}
And then I have no idea how to do the rest, and og What I’m doing is right
And what about $P(X > m+l)$ how do I find that ?
Part 2: any hints?
Best Answer
You made a mistake when computing $p\sum_0^\infty (1-p)^{i-1}$. It should be $\frac1{1-(1-p)} = 1/p$.
If you wanted to continue the way you are going, you could use the formula for a finite geometric series: $$ \sum_{0}^m x^n = \frac{1-x^{m+1}}{1-x} $$ However, an easier way is to recognize that the tail sum $\sum_{i=m+1}^\infty P(X=i)$ is a geometric series, so you can directly apply the infinite geometric series formula to it: $$ P(X>m) = \sum_{i=m+1}^\infty P(X=i) = p\sum_{i=m+1}^\infty (1-p)^{i-1} = p\frac{(1-p)^m}{1-(1-p)} = (1-p)^m $$ This allows you to compute $P(X>m+\ell),P(X>\ell)$, and therefore $P(X>m+\ell|X>m)$.
For (2), let $\alpha = P(X>1)$. Using the given relation (plug in $\ell=1$), you can show that $$ P(X>m+1) = \alpha P(X>m) $$ Use this to prove $P(X>m) = \alpha^m$ by induction, and then use this to compute $P(X=m)$.